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algebra

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the dimensions of a rectangle are such that its length is 5 in more than its width. If the length were doubled and the width were decreases by 2 in, the area would be increased 136 in^2,what are the length and width of the rectangle

  • algebra - ,

    Initially
    w=width
    w+5=length
    w(w+5)=original area

    Subsequently:
    w-2=width
    2(w+5)=length
    (w-2)*2(w+5)=2*original area=w(w+5)=area

    Therefore
    2(w-2)(w+5)=w(w+5)

    Solve for w:
    2(w-2)=w (w cannot be equal to -5)
    2w-4=w
    w=4

  • algebra - ,

    original width -- x
    original length -- x+5
    area = x(x+5)

    new width = x-2
    new length = 2x+10
    new area = (2x+10)(x-2)

    (2x+10)(x-2) - x(x+5) = 136
    2x^2 + 6x - 20 - x^2 - 5x - 136 = 0
    x^2 + x - 156 = 0
    (x-12)(x+13) = 0
    x = 12 or x = -13

    ignoring the negative,

    width = 12
    length = 17


    check:
    original area = 12(17) = 204

    new width = 10
    new length = 34
    new area = 10(34) = 340

    increase in area = 340-204 = 136 , YEahh!

  • algebra - ,

    Good catch!
    Thank you!

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